Irodov 3.218: Avalanches Started Everywhere at Once – Amit & Deepali Sinha Foundation

Problem Irodov 3.218

Ultraviolet radiation ionizes the gas between the plates of a capacitor, a distance $d$ apart, uniformly, forming $n_i$ electrons per unit volume per second. Driven by the field of the capacitor, these electrons ionize the gas molecules, each electron producing $\alpha$ new electrons (and as many ions) per unit length of its path. Ionization by the ions is negligible. Find the electron current density at the plate at the higher potential.

The gas-filled capacitor, its plates a distance d apart. — The gas-filled capacitor, its plates a distance $d$ apart.

Given

gap $d$
$n_i$ electrons formed per unit volume per second, uniformly
$\alpha$ new electrons per unit path length
no ionization by ions
steady state, no electrons emitted by the plates, none recombine (assumed)

Find

the electron current density

j

at the higher-potential plate

Solution

Step 1The balance in a thin layer

The electrons drift against the field, from the plate at the lower potential to the plate at the higher. Here $n_i$ is a rate, the number of electrons the radiation forms per unit volume per second, not a concentration. Measure $x$ from the lower-potential plate, and let $\Gamma(x)$ electrons cross unit area of the plane $x$ per second in the steady state. The layer between $x$ and $x + dx$ adds the electrons the radiation forms in it and those the crossing electrons knock out:

d\Gamma = n_i\,dx + \alpha\,\Gamma\,dx

No electron enters the gap at the lower-potential plate (assumed: the plates emit none), so $\Gamma(0) = 0$ .

Step 2Solving the linear equation

Written as $d\Gamma/dx - \alpha\Gamma = n_i$ , the equation becomes an exact derivative on multiplying by $e^{-\alpha x}$ :

\frac{d}{dx}\left(\Gamma e^{-\alpha x}\right) = n_i\,e^{-\alpha x}

Integrating from $0$ , where $\Gamma = 0$ , to $x$ :

\Gamma e^{-\alpha x} = n_i\int_0^x e^{-\alpha x'}\,dx' = \frac{n_i}{\alpha}\left(1 - e^{-\alpha x}\right)

Multiplying through by $e^{\alpha x}$ :

\Gamma(x) = \frac{n_i}{\alpha}\left(e^{\alpha x} - 1\right) \tag{1}

The electron current density across the gap, drawn for α d = 1.2, from zero at the lower-potential plate to its largest value at the plate at x = d. Dashed: the same ionization with no multiplication, growing only linearly. — The electron current density across the gap, drawn for $\alpha d = 1.2$ , from zero at the lower-potential plate to its largest value at the plate at $x = d$ . Dashed: the same ionization with no multiplication, growing only linearly.

Step 3The current density at the higher-potential plate

Each electron carries the charge $e$ , so the electron current density at $x = d$ is $e$ times (1) there:

j = \frac{e\,n_i}{\alpha}\left(e^{\alpha d} - 1\right)

(2)

Check

Limit: for $\alpha d \ll 1$ , $e^{\alpha d} - 1 \approx \alpha d$ and $j \to en_i d$ , the saturation current density of a gap whose electrons are only collected, all those formed in a column of unit area. Second route, by superposition: an electron formed at $x'$ grows into $e^{\alpha(d - x')}$ electrons by the time it arrives (as in a single avalanche), so

j = e\int_0^d n_i\,e^{\alpha(d - x')}\,dx' = \frac{e\,n_i}{\alpha}\left(e^{\alpha d} - 1\right),

the boxed result.

Remark

Unless $\alpha d$ is small, most of the current comes from electrons formed near the lower-potential plate, which have the whole gap to multiply in: the layer next to it contributes $e^{\alpha d}$ times as much per unit thickness as the layer next to the collecting plate. The gain over plain collection, $(e^{\alpha d} - 1)/(\alpha d)$ , is smaller than the $e^{\alpha d}$ of an avalanche started at the cathode, because on average the electrons travel only part of the gap.